TITLE: Does $a_n + a_{n+1} + a_{n+2} \to 0$ imply $a_n \to 0$?
QUESTION [7 upvotes]: Given $\{{a_n}\}_{n=0}^{\infty}$ and $\{{b_n}\}_{n=0}^{\infty}$
Prove or disprove:
1) if $\lim \limits_{n\to \infty}a_n=0$ then $\lim \limits_{n\to \infty}a_n-[a_n]=0$
I think (1) is correct because if $\lim \limits_{n\to \infty}a_n=0$ then by the definition of limit I can show that for each $\epsilon>0, |a_n-[a_n]|<\epsilon$ 
2)If $a_n$ converges and $b_n$ doesn't converge then $(a_n+b_n)$ doesn't converge.
I think (2) is correct, but I'm not sure how to start proving it - maybe I can assume that it isn't correct and then get a contradiction?
3)If $\lim \limits_{n\to \infty}\frac{a_n+a_{n+1}+a_{n+2}}{3}=0$ then $\lim \limits_{n\to \infty}a_n=0$
I have no idea about (3).
My knowledge is of simple calculus theorem(limit definition, arithmetics of limits and the Squeeze Theorem).
Thanks a lot for your time and help.

REPLY [12 votes]: The assertion is not true. Indeed, $a_n$ converges to $0$, so we have to see whether $[a_n]$ (I think it's the floor function) converges to $0$. But it doesn't need to be the case, for example with $a_n=-\frac 1n$.
You can show it's true by contradiction: if $a_n+b_n$ converges, since $-a_n$ converges and the sum of two converging sequences is... 
The sequence $a_n=\left(\frac{1+i\sqrt 3}2\right)^n$ is such that $a_n+a_{n+1}+a_{n+2}=0$ but doesn't converge to $0$. Alternatively, a simpler example given by @robjohn is $a_k:=2\cos\left(\frac{2\pi}3k\right)$.

REPLY [10 votes]: 1) What is $|a_n - [a_n]|$, if $|a_n - 0| < \epsilon < \frac{1}{2}$?
2) Let $a = \lim_{n \to \infty}a_n$, then $|a_n + b_n - x| \geq |b_n - x + a| - |a_n - a| \geq |b_n - x'| - \epsilon$ for $x' = x-a$ and $n$ big enough.
3) $1,-1,0,1,-1,0,\dots$